$lim _{xrightarrow c}f(x)=L$ if and only if $lim_{xrightarrow0}f(x+c)=L$












4












$begingroup$



Let $f:= mathbb{R}tomathbb{R}$ and let $cinmathbb{R}$. Show that $lim _{xrightarrow c}f(x)=L$ if and only if $lim_{xrightarrow0}f(x+c)=L$.




From the definition of limit, we get that it is enough to show:

$forall$ $varepsilon>0$ $exists$ $delta>0$ s.t. if $|x-c|<delta$ then $|f(x)-L|<varepsilon$
$Leftrightarrow$ $forall$ $varepsilon_0>0$ $exists$ $delta_0>0$ s.t. if $|x|<delta_0$ then $|f(x+c)-L|<varepsilon_0$




I can replace $x$ by $x+c$ everywhere in statement for the if $(Rightarrow)$ part. But, I am not sure this is the correct method. What I need to do is manipulate the inequalities in each to get the other. But, I am not sure how to proceed with that.






real-analysis limits analysis







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    You mean $lim_{x to c} f(x) = L$ if and only if $lim_{x to 0} f(x+c) = L$.
    $endgroup$
    – Robert Israel
    Dec 19 '18 at 3:03










  • $begingroup$
    @RobertIsrael corrected!
    $endgroup$
    – Za Ira
    Dec 19 '18 at 3:07
















4












$begingroup$



Let $f:= mathbb{R}tomathbb{R}$ and let $cinmathbb{R}$. Show that $lim _{xrightarrow c}f(x)=L$ if and only if $lim_{xrightarrow0}f(x+c)=L$.




From the definition of limit, we get that it is enough to show:

$forall$ $varepsilon>0$ $exists$ $delta>0$ s.t. if $|x-c|<delta$ then $|f(x)-L|<varepsilon$
$Leftrightarrow$ $forall$ $varepsilon_0>0$ $exists$ $delta_0>0$ s.t. if $|x|<delta_0$ then $|f(x+c)-L|<varepsilon_0$




I can replace $x$ by $x+c$ everywhere in statement for the if $(Rightarrow)$ part. But, I am not sure this is the correct method. What I need to do is manipulate the inequalities in each to get the other. But, I am not sure how to proceed with that.






real-analysis limits analysis







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    You mean $lim_{x to c} f(x) = L$ if and only if $lim_{x to 0} f(x+c) = L$.
    $endgroup$
    – Robert Israel
    Dec 19 '18 at 3:03










  • $begingroup$
    @RobertIsrael corrected!
    $endgroup$
    – Za Ira
    Dec 19 '18 at 3:07














4












4








4


1



$begingroup$



Let $f:= mathbb{R}tomathbb{R}$ and let $cinmathbb{R}$. Show that $lim _{xrightarrow c}f(x)=L$ if and only if $lim_{xrightarrow0}f(x+c)=L$.




From the definition of limit, we get that it is enough to show:

$forall$ $varepsilon>0$ $exists$ $delta>0$ s.t. if $|x-c|<delta$ then $|f(x)-L|<varepsilon$
$Leftrightarrow$ $forall$ $varepsilon_0>0$ $exists$ $delta_0>0$ s.t. if $|x|<delta_0$ then $|f(x+c)-L|<varepsilon_0$




I can replace $x$ by $x+c$ everywhere in statement for the if $(Rightarrow)$ part. But, I am not sure this is the correct method. What I need to do is manipulate the inequalities in each to get the other. But, I am not sure how to proceed with that.






real-analysis limits analysis







share|cite|improve this question











$endgroup$





Let $f:= mathbb{R}tomathbb{R}$ and let $cinmathbb{R}$. Show that $lim _{xrightarrow c}f(x)=L$ if and only if $lim_{xrightarrow0}f(x+c)=L$.




From the definition of limit, we get that it is enough to show:

$forall$ $varepsilon>0$ $exists$ $delta>0$ s.t. if $|x-c|<delta$ then $|f(x)-L|<varepsilon$
$Leftrightarrow$ $forall$ $varepsilon_0>0$ $exists$ $delta_0>0$ s.t. if $|x|<delta_0$ then $|f(x+c)-L|<varepsilon_0$




I can replace $x$ by $x+c$ everywhere in statement for the if $(Rightarrow)$ part. But, I am not sure this is the correct method. What I need to do is manipulate the inequalities in each to get the other. But, I am not sure how to proceed with that.






real-analysis limits analysis




real-analysis limits analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 3:39









Martin Argerami

129k1184185




129k1184185










asked Dec 19 '18 at 2:57









Za IraZa Ira

161115




161115








  • 4




    $begingroup$
    You mean $lim_{x to c} f(x) = L$ if and only if $lim_{x to 0} f(x+c) = L$.
    $endgroup$
    – Robert Israel
    Dec 19 '18 at 3:03










  • $begingroup$
    @RobertIsrael corrected!
    $endgroup$
    – Za Ira
    Dec 19 '18 at 3:07














  • 4




    $begingroup$
    You mean $lim_{x to c} f(x) = L$ if and only if $lim_{x to 0} f(x+c) = L$.
    $endgroup$
    – Robert Israel
    Dec 19 '18 at 3:03










  • $begingroup$
    @RobertIsrael corrected!
    $endgroup$
    – Za Ira
    Dec 19 '18 at 3:07








4




4




$begingroup$
You mean $lim_{x to c} f(x) = L$ if and only if $lim_{x to 0} f(x+c) = L$.
$endgroup$
– Robert Israel
Dec 19 '18 at 3:03




$begingroup$
You mean $lim_{x to c} f(x) = L$ if and only if $lim_{x to 0} f(x+c) = L$.
$endgroup$
– Robert Israel
Dec 19 '18 at 3:03












$begingroup$
@RobertIsrael corrected!
$endgroup$
– Za Ira
Dec 19 '18 at 3:07




$begingroup$
@RobertIsrael corrected!
$endgroup$
– Za Ira
Dec 19 '18 at 3:07










1 Answer
1






active

oldest

votes


















4












$begingroup$

Given $varepsilon>0$, use the $delta$ from the definition of $lim_{xto c}f(x)=L$. If $|x|<delta$, then $|(x+c)-c|<delta$, so $|f(x+c)-L|<varepsilon$. Thus $lim_{xto0}f(x+c)=L$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    so, my approach was correct?
    $endgroup$
    – Za Ira
    Dec 19 '18 at 3:41






  • 1




    $begingroup$
    I personally wouldn't be comfortable with "replacing". Your replacement occurs within two quantifiers and inside an implication. I know it works, but if I try to do it your way, it is not easy for me to justify it.
    $endgroup$
    – Martin Argerami
    Dec 19 '18 at 3:43












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Given $varepsilon>0$, use the $delta$ from the definition of $lim_{xto c}f(x)=L$. If $|x|<delta$, then $|(x+c)-c|<delta$, so $|f(x+c)-L|<varepsilon$. Thus $lim_{xto0}f(x+c)=L$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    so, my approach was correct?
    $endgroup$
    – Za Ira
    Dec 19 '18 at 3:41






  • 1




    $begingroup$
    I personally wouldn't be comfortable with "replacing". Your replacement occurs within two quantifiers and inside an implication. I know it works, but if I try to do it your way, it is not easy for me to justify it.
    $endgroup$
    – Martin Argerami
    Dec 19 '18 at 3:43
















4












$begingroup$

Given $varepsilon>0$, use the $delta$ from the definition of $lim_{xto c}f(x)=L$. If $|x|<delta$, then $|(x+c)-c|<delta$, so $|f(x+c)-L|<varepsilon$. Thus $lim_{xto0}f(x+c)=L$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    so, my approach was correct?
    $endgroup$
    – Za Ira
    Dec 19 '18 at 3:41






  • 1




    $begingroup$
    I personally wouldn't be comfortable with "replacing". Your replacement occurs within two quantifiers and inside an implication. I know it works, but if I try to do it your way, it is not easy for me to justify it.
    $endgroup$
    – Martin Argerami
    Dec 19 '18 at 3:43














4












4








4





$begingroup$

Given $varepsilon>0$, use the $delta$ from the definition of $lim_{xto c}f(x)=L$. If $|x|<delta$, then $|(x+c)-c|<delta$, so $|f(x+c)-L|<varepsilon$. Thus $lim_{xto0}f(x+c)=L$.






share|cite|improve this answer









$endgroup$



Given $varepsilon>0$, use the $delta$ from the definition of $lim_{xto c}f(x)=L$. If $|x|<delta$, then $|(x+c)-c|<delta$, so $|f(x+c)-L|<varepsilon$. Thus $lim_{xto0}f(x+c)=L$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 3:13









Martin ArgeramiMartin Argerami

129k1184185




129k1184185












  • $begingroup$
    so, my approach was correct?
    $endgroup$
    – Za Ira
    Dec 19 '18 at 3:41






  • 1




    $begingroup$
    I personally wouldn't be comfortable with "replacing". Your replacement occurs within two quantifiers and inside an implication. I know it works, but if I try to do it your way, it is not easy for me to justify it.
    $endgroup$
    – Martin Argerami
    Dec 19 '18 at 3:43


















  • $begingroup$
    so, my approach was correct?
    $endgroup$
    – Za Ira
    Dec 19 '18 at 3:41






  • 1




    $begingroup$
    I personally wouldn't be comfortable with "replacing". Your replacement occurs within two quantifiers and inside an implication. I know it works, but if I try to do it your way, it is not easy for me to justify it.
    $endgroup$
    – Martin Argerami
    Dec 19 '18 at 3:43
















$begingroup$
so, my approach was correct?
$endgroup$
– Za Ira
Dec 19 '18 at 3:41




$begingroup$
so, my approach was correct?
$endgroup$
– Za Ira
Dec 19 '18 at 3:41




1




1




$begingroup$
I personally wouldn't be comfortable with "replacing". Your replacement occurs within two quantifiers and inside an implication. I know it works, but if I try to do it your way, it is not easy for me to justify it.
$endgroup$
– Martin Argerami
Dec 19 '18 at 3:43




$begingroup$
I personally wouldn't be comfortable with "replacing". Your replacement occurs within two quantifiers and inside an implication. I know it works, but if I try to do it your way, it is not easy for me to justify it.
$endgroup$
– Martin Argerami
Dec 19 '18 at 3:43


















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